N-th Tribonacci Number Leetcode Problem 1137

Are you ready to dive into another coding challenge?

Today, we'll tackle the "N-th Tribonacci Number" problem.

This is an interesting challenge for beginners, and by the end of this blog post, you'll have a solid understanding of how to approach it.

We'll explore both the brute-force and efficient approaches and explain the reasoning behind the efficient solution.

Before we start, you can find the problem description and practice it on LeetCode by following this link.

Problem Overview

The Tribonacci sequence, denoted as Tn, is defined as follows:
– T0 = 0
– T1 = 1
– T2 = 1
– Tn+3 = Tn + Tn+1 + Tn+2 for n >= 0

In simpler terms, each number in the Tribonacci sequence is the sum of the three previous numbers.

For example, T3 = 0 + 1 + 1 = 2, T4 = 1 + 1 + 2 = 4, and so on.

Our task is to find the value of the N-th Tribonacci number, given an integer N.

Understanding the Constraints

Before we dive into the solution, it's crucial to understand the problem constraints.

In this case, we have the following constraints:

0 <= n <= 37
The answer is guaranteed to fit within a 32-bit integer, i.e., answer <= 2^31 – 1. These constraints tell us that we'll be working with relatively small values of N, and we don't need to worry about integer overflow.

Brute-force Approach

Let's start with the brute-force approach.

The idea here is to compute each Tribonacci number iteratively from the beginning up to the N-th number.

This approach is straightforward but not the most efficient.

def tribonacci(n: int) -> int:
    if n < 2:
        return n  # Base cases for T0 and T1
    elif n == 2:
        return 1  # Base case for T2
    
    t0, t1, t2 = 0, 1, 1
    
    for i in range(3, n + 1):
        t3 = t0 + t1 + t2
        t0, t1, t2 = t1, t2, t3
    
    return t2

In the brute-force approach, we initialize three variables (t0, t1, and t2) to represent the previous three Tribonacci numbers.

We start a loop from 3 to N and compute each new Tribonacci number based on the previous three values.

Finally, we return t2, which holds the N-th Tribonacci number.

The time complexity of this approach is O(n) because we need to calculate the Tribonacci numbers up to the N-th number.

The space complexity is O(1) because we only use a constant amount of extra space.

Efficient Approach with Memoization

The brute-force approach works fine for small values of N.

However, there's an even more efficient way to solve this problem using memoization.

By storing previously computed Tribonacci numbers in memory, we can avoid redundant calculations and significantly speed up the process.

Let's implement the efficient solution:

View on GitHub ➹

class Solution:
    memo = {}

    def tribonacci(self, n: int) -&gt; int:
        if n in self.memo:
            return self.memo[n]
        if n == 0:
            return 0
        if n == 1:
            return 1
        if n == 2:
            return 1
        self.memo[n] = (
            self.tribonacci(n - 1) + self.tribonacci(n - 2) + self.tribonacci(n - 3)
        )
        return self.memo[n]

In the efficient approach, we use a dictionary (memo) to store previously computed Tribonacci numbers.

If the N-th number is already in the memo, we simply return it, avoiding any extra computation.

If N is one of the base cases (0, 1, or 2), we return the corresponding value.

Otherwise, we calculate the N-th Tribonacci number as the sum of the three previous numbers using recursion.

The result is stored in the memo dictionary for future reference.

This approach significantly reduces the number of calculations and is much faster, especially for larger values of N.

Time and Space Complexity

Let's analyze the time and space complexity of our efficient solution:

Time Complexity: The time complexity of this approach is O(n), which is the number of unique calculations we need to make to compute the N-th Tribonacci number.

However, thanks to memoization, we avoid redundant calculations, making this approach much faster in practice.

Space Complexity: The space complexity is O(n), mainly due to the memo dictionary.

As we store the results of each unique calculation, the space required grows linearly with the input N.

Reasoning Behind Our Approach

The reasoning behind our efficient approach is to avoid redundant calculations by using memoization.

We know that each Tribonacci number depends on the three previous numbers, so we store these values and use them to calculate the N-th number.

This technique optimizes the solution and ensures that we don't waste time recalculating values we've already determined.

Conclusion

In this blog post, we tackled the "N-th Tribonacci Number" problem on LeetCode.

We explored two approaches: the brute-force approach, which calculates each Tribonacci number iteratively, and the efficient approach with memoization, which significantly improves the algorithm's speed.

Understanding the problem constraints, implementing these approaches, and reasoning about their efficiency are essential skills for any coder.

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If you have questions, suggestions, or would like to share your thoughts, please leave a comment.

Happy coding!